Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $47.2$ years; the standard deviation is $4.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $42.6$ and $51.8$ years.
Answer: $47.2$ $42.6$ $51.8$ $38$ $56.4$ $33.4$ $61$ $68\%$ We know the lifespans are normally distributed with an average lifespan of $47.2$ years. We know the standard deviation is $4.6$ years, so one standard deviation below the mean is $42.6$ years and one standard deviation above the mean is $51.8$ years. Two standard deviations below the mean is $38$ years and two standard deviations above the mean is $56.4$ years. Three standard deviations below the mean is $33.4$ years and three standard deviations above the mean is $61$ years. We are interested in the probability of a bear living between $42.6$ and $51.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the bears will have lifespans within 1 standard deviation of the average lifespan. The probability of a particular bear living between $42.6$ and $51.8$ years is ${68\%}$.